Підсумки про оптимізацію на відкритих чи необмежених інтервалах

Optimization on open or unbounded intervals is almost exactly the same as optimization on closed intervals.

Suppose that we have a function $f(x)$ which we would like to optimize (maximize, minimize, take your pick) over some open or unbounded interval.

The first step is to find the critical numbers. These are all numbers $c$ such that $f'(c)=0$ or $f'(c)$ is undefined.

Now, to find the relative maxima and minima, one may use any of the tests that we are familiar with, such as simply charting the values of $f'(x)$ to see where it is positive or negative, or the second derivative test.

However, we are interested in the absolute extrema, and this is going to require simply checking the values in the general case. Check the value of $f(x)$ at all critical numbers plus the boundary cases! If the function is unbounded on one or both sides, check the value of the function "at infinity." That is, compute $\lim_{x\to\infty}f(x)$ or $\lim_{x\to-\infty}f(x)$.

Out of all of these values, you should find a maximum and a minimum value. If one or both of these values are at the boundaries (or at infinity), you should discard them and say that there is no maximum/minimum value. This is because the interval is open, so it does not contain the boundary. However, if the extreme value is one of the critical points, you can declare than as the optimum value on the interval.

Here is an example: compute the extreme values of $f(x)=x^3-3x$ on the open interval $(-2,3)$.

First, we find the critical points. Since $f(x)$ is a polynomial, the derivative is always defined: $$f'(x)=0$$ $$\frac{d}{dx}(x^3-3x)=0$$ $$3x^2-3=0$$ $$x=1,-1$$ Compute the value of $f(x)$ at the critical points and at the boundaries: $$f(-2)=-2$$ $$f(-1)=2$$ $$f(1)=-2$$ $$f(3)=18$$ The maximum value is $18$, at $x=3$. But this is a boundary value, so we discard it. Thus, there is no maximum value on this interval, although we may obtain values arbitrarily close to $18$.

The minimum value is $-2$, at $x=-2,1$. We discard $x=-2$ because it is a boundary, but we are still left with one value of $x$ at which $f(x)$ achieves its minimum. Thus, the minimum value is $-2$, at $x=1$.

In other words, trivial inequality $x^2 \ge 0$ is trivial.

That's it! Many times you can solve optimization problems just by thinking a bit, instead of mechanically using calculus. :)

Now let's try a similar problem:

Yay! It's the same idea as before. The square of a real number is always at least $0$, so $(x^2+6x+2)^2 \geq 0$ for any real $x$. Furthermore, if $x^2+6x+2 = (x+3)^2 - 7 = 0$ (for the real roots $x = 3 \pm \sqrt{7}$), we get $(x^2+6x+2)^2 = 0^2 = 0$. So the minimum is just $0$.

If you'd like to see more of the HMMT November 2013 Guts Round, click here.